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Klio2033 [76]
2 years ago
11

A dot plot titled Distance from School in Blocks going from 1 to 6. 1 has 6 dots, 2 has 5 dots, 3 has 4 dots, 4 has 2 dots, 5 ha

s 3 dots, and 6 has 2 dots.
Wynn needs to find the center of the data set shown on the dot plot.

The dot plot has
dots.

The dot plot has
dots to the left of the
center and
dots to the right of the center.

The center of the data set is
.
Mathematics
2 answers:
Basile [38]2 years ago
7 0

Answer:

Sorry its late

Step-by-step explanation:

A dot plot titled Distance from School in Blocks going from 1 to 6. 1 has 6 dots, 2 has 5 dots, 3 has 4 dots, 4 has 2 dots, 5 has 3 dots, and 6 has 2 dots.

Wynn needs to find the center of the data set shown on the dot plot.

The dot plot has

✔ 22

dots.

The dot plot has

✔ 11

dots to the left of the

center and

✔ 11

dots to the right of the center.

The center of the data set is

✔ between 2 and 3

.

Burka [1]2 years ago
3 0
<h2>22,11,11, between 2 and 3</h2>

Edge 2022

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A bucket that weighs 4 lb and a rope of negligible weight are used to draw water from a well that is 60 ft deep. The bucket is f
jonny [76]

Answer:

2580 ft-lb

Step-by-step explanation:

Water leaks out of the bucket at a rate of \frac{0.15 \mathrm{lb} / \mathrm{s}}{1.5 \mathrm{ft} / \mathrm{s}}=0.1 \mathrm{lb} / \mathrm{ft}

Work done required to pull the bucket to the top of the well is given by integral

W=\int_{a}^{b} F(x) dx

Here, function F(x) is the total weight of the bucket and water x feet above the bottom of the well. That is,

F(x)=4+(42-0.1 x)

=46-0.1x

a is the initial height and b is the maximum height of well. That is,

a=0 \text { and } b=60

Find the work done as,

W=\int_{a}^{b} F(x) d x

=\int_{0}^{60}(46-0.1 x) dx

&\left.=46x-0.05 x^{2}\right]_{0}^{60}

=(2760-180)-0[

=2580 \mathrm{ft}-\mathrm{lb}&#10;

Hence, the work done required to pull the bucket to the top of the well is 2580 \mathrm{ft}- \mathrm{lb}

4 0
3 years ago
A farmer wishes to enclose a pasture that is bordered on one side by a river (so one of the four sides wont require fencing) She
docker41 [41]
The area enclosed is (2x)(600-2x).(2x)(600−2x)=1200x−4x2
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1200=8x
x=150
Therefore the sides for maximum area are 150*300.
<span>The area is: 45000</span>
6 0
3 years ago
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[20points][Precal] Find the cross product of -(3/4)V and (1/2)w if v = &lt;-2, 12, -3&gt; and w = &lt;-7, 4, -6&gt;
Lana71 [14]
\mathbf v=\langle-2,12,-3\rangle=-2\,\vec i+12\,\vec j-3\,\vec k
-\dfrac34\mathbf v=\dfrac32\,\vec i-9\,\vec j+\dfrac94\,\vec k

\mathbf w=\langle-7,4,-6\rangle=-7\,\vec i+4\,\vec j-6\,\vec k
\dfrac12\mathbf w=-\dfrac72\,\vec i+2\,\vec j-3\,\vec k

-\dfrac34\mathbf v\times\dfrac12\mathbf w=\begin{vmatrix}\vec i&\vec j&\vec k\\\frac32&-9&\frac94\\-\frac72&2&-3\end{vmatrix}
=\left((-9)\times(-3)-\dfrac94\times2\right)\,\vec i-\left(\dfrac32\times(-3)-\dfrac94\times\left(-\dfrac72\right)\right)\,\vec j+\left(\dfrac32\times2-(-9)\times\left(-\dfrac72\right)\right)\,\vec k
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6 0
3 years ago
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Mazyrski [523]

Step-by-step explanation:

1. d = rt

Divide t from both sides.

\frac{d}{t} = r

2. PV = nRT

Divide P from both sides.

V = \frac{nRT}{P}

3. \frac{x}{2} -g=a\\

Add g to both sides.

\frac{x}{2} = g + a

Now multiply by 2 on both sides.

x = 2(g + a)

(or, x = 2g + 2a)

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Divide g from both sides.

h + 1.5 = \frac{11}{g}

Now subtract 1.5 from both sides.

h = \frac{11}{g} -1.5

5. a(n-2)+5=bn\\

n = \frac{2a-5}{a - b}

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pashok25 [27]

Answer:

D i got u girl

Step-by-step explanation:

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